Chapter 3: Problem 21
In Problems \(15-24\), find the values of \(x \in \mathbf{R}\) for which the given functions are continuous. $$ f(x)=\ln \frac{x}{x+1} $$
Short Answer
Expert verified
The function is continuous for \(x \in (-\infty, -1) \cup (0, \infty)\).
Step by step solution
01
Identify Conditions for Continuity.
To determine where the function \(f(x) = \ln \frac{x}{x+1}\) is continuous, we first need to recognize that for the natural logarithm \(\ln(u)\) to be defined and continuous, \(u\) itself must be positive, \(u > 0\). Additionally, for \(\ln(x)\), \(x > 0\).
02
Set the Argument Condition
The argument of the logarithm function \(u(x) = \frac{x}{x+1}\) must be greater than zero. Thus, we set the inequality: \[ \frac{x}{x+1} > 0 \] This requires solving the inequality for \(x\).
03
Solve the Inequality
To solve \(\frac{x}{x+1} > 0\), consider the critical points that make the numerator \(x\) and the denominator \(x+1\) equal to zero: - \(x = 0\) makes the numerator zero.- \(x = -1\) makes the denominator undefined.Analysis of the intervals determined by these points gives:- \((-\infty, -1)\): both numerator and denominator are negative, so the fraction is positive.- \((-1, 0)\): numerator is negative and the denominator is positive, making the fraction negative.- \((0, +\infty)\): both numerator and denominator are positive, so the fraction is positive.Thus, the function is defined, and the inequality holds in the intervals \((-\infty, -1) \cup (0, \infty)\).
04
Analyze Continuity in Given Intervals
There are no discontinuities within the intervals \((-\infty, -1)\) and \((0, \infty)\) as both the logarithm and rational functions are continuous where they are defined. The critical point \(x = -1\) is where the function is undefined, creating discontinuity, and at \(x = 0\), \(f(x) = \ln(0)\) is undefined as well.
05
Conclude the Solution
The function is continuous on the domain \(x \in (-\infty, -1) \cup (0, \infty)\). This includes regions where the natural log function is defined and the rational expression is greater than zero, avoiding points of undefined behavior.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm Functions
The natural logarithm function, often written as \ln(x)\, is a fundamental mathematical function with a wide range of applications. At its core, a natural logarithm answers the question: "To what power must the base ***e*** be raised to get ***x***?" Here, ***e*** is a mathematical constant approximately equal to 2.71828.
A critical point to remember about natural logarithms is that they are only defined for positive values. In other words, the argument of a natural logarithm must always be positive. This implies \( x > 0 \) for the function \( \ln(x) \) to exist. In scenarios involving more complex expressions, such as \( \ln \left( \frac{x}{x+1} \right) \), the entire expression \( \frac{x}{x+1} \) should be greater than zero to ensure the logarithm is defined.
Natural logarithms have properties that can simplify various algebraic manipulations such as:
A critical point to remember about natural logarithms is that they are only defined for positive values. In other words, the argument of a natural logarithm must always be positive. This implies \( x > 0 \) for the function \( \ln(x) \) to exist. In scenarios involving more complex expressions, such as \( \ln \left( \frac{x}{x+1} \right) \), the entire expression \( \frac{x}{x+1} \) should be greater than zero to ensure the logarithm is defined.
Natural logarithms have properties that can simplify various algebraic manipulations such as:
- \( \ln(ab) = \ln(a) + \ln(b) \) - the logarithm of a product.
- \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \) - the logarithm of a quotient.
- \( \ln(a^b) = b \cdot \ln(a) \) - the logarithm of a power.
Solving Inequalities
Inequalities are mathematical expressions involving relations like greater than (>), less than (<), greater than or equal to (≥), and less than or equal to (≤). Solving an inequality can sometimes be like solving equations, but with some important differences.
When dealing with inequalities, particularly those involving fractions or rational expressions like \( \frac{x}{x+1} \), it's essential to consider the critical points that make either the numerator or denominator zero. Identifying these points helps break down the inequality into intervals.
For \( \frac{x}{x+1} > 0 \), you would:
When dealing with inequalities, particularly those involving fractions or rational expressions like \( \frac{x}{x+1} \), it's essential to consider the critical points that make either the numerator or denominator zero. Identifying these points helps break down the inequality into intervals.
For \( \frac{x}{x+1} > 0 \), you would:
- Identify the zeros: \( x = 0 \) (numerator) and the undefined point \( x = -1 \) (denominator).
- Analyze the signs over the intervals created by these critical points: \((-\infty, -1)\), \((-1, 0)\), and \((0, \infty)\). This involves checking if the entire expression is positive or negative within each interval.
Domain of a Function
The domain of a function is a set of all possible input values (or x-values) for which the function is defined. Understanding the domain is crucial as it delineates where a function can safely be evaluated without running into undefined operations like division by zero or taking the logarithm of a non-positive value.
For the function \( f(x) = \ln \frac{x}{x+1} \), the domain is determined by ensuring the expression inside the logarithm stays positive and avoids division by zero. Thus, you first find where \( \frac{x}{x+1} > 0 \), carefully checking critical points like \( x = -1 \) where the expression becomes undefined.
The intervals \((-\infty, -1) \cup (0, \infty)\) define the domain in this case. This indicates:
For the function \( f(x) = \ln \frac{x}{x+1} \), the domain is determined by ensuring the expression inside the logarithm stays positive and avoids division by zero. Thus, you first find where \( \frac{x}{x+1} > 0 \), carefully checking critical points like \( x = -1 \) where the expression becomes undefined.
The intervals \((-\infty, -1) \cup (0, \infty)\) define the domain in this case. This indicates:
- For \( x \in (-\infty, -1) \), both the numerator and the denominator are negative, keeping \( \frac{x}{x+1} \) positive.
- For \( x \in (0, \infty) \), both the numerator and the denominator are positive, also resulting in a positive quotient.